Question number 19. Which of the graphs below shows the solution set of the following equation? So this equation here I will not pronounce for you, but to decide which graphs are possible, I will pick coordinates on each of the graphs here, trying to double up as much as possible on these coordinates, and then see if that point does indeed belong to the solution set. So here I have an A, the y-intercept 0, negative 1, and then an E, also the y-intercept 0, negative 1. So 0, negative 1 here gives us, if x is equal to 0, I have 0 quantity squared, plus if y is equal to negative 1, negative 1 plus the absolute value of negative 1 quantity squared, which is equal to negative 2 squared, and that is equal to 4. So that works. So these points here do belong to the solution set, so let me just move on for now to the other points. I have in B coordinates 1, 0, the x- intercept. So 1, 0 produces for us the value on the left-hand side, I have 1 minus the absolute value of 1 quantity squared. If y is 0, then I have plus 0 squared, and so that's 0 plus 0, and that is not equal to 4. So here, choice B, we discount right away, and then in C and in D, I have coordinates 2, 0 for those x-intercepts, so I can deal with them at the same time, and we then see that 2, 0 gives, on the left-hand side, if x is 2, 2 minus the absolute value of 2 quantity squared, plus if y is 0, 0 squared, and again that's 0 plus 0, which is not equal to 4. So choices C and D can be discounted. Now we also note that between E and A, the remaining two possibilities, I have both the same x-intercepts, negative 1, 0, and negative 1, 0, so there is no point in trying to use those, but what we will do is then see, is the graph here between those intercepts more like a line, which is the behavior in E, or like a circle, which is the behavior in A. So let now x and y be negative, so the points lie in quadrant 3, and then we simplify. So if x is negative, the absolute value of x is equal to the negative of the absolute value of x, and if y is negative, then the absolute value of y is equal to the negative of the absolute value of y, like that, and we use that to simplify, so we have in our equation here. So the equation then looks like what we have is x minus and negative x quantity squared plus y minus negative y quantity squared, and that's equal to 4. So we can simplify here. We have 2x quantity squared plus 2y quantity squared equal to 4. So that's 4x squared plus 4y squared equal to 4. And then dividing by 4 on both sides, we have x squared plus y squared is equal to 1. And this thing here is a equation for the unit circle. So the graph here in quadrant 3 behaves like a circle, and that is clearly what happens in graph A. So that's what we choose as our answer here. So answer A is the correct solution.

unit circle

graph solution

mathematical equation

coordinates testing

intercepts calculation